The sum of the non-real root of `(x^2+x-2)(x^2+x-3)=12` is `-1` b. `1` c. `-6` d. `6`

Options

A. -1
B. 1
C. -6
D. 6

Explanation

To solve the equation \((x^2 + x - 2)(x^2 + x - 3) = 12\) and find the sum of the non-real roots, we can follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the equation:
\[ (x^2 + x - 2)(x^2 + x - 3) = 12 \]
This can be rearranged as:
\[ (x^2 + x - 2)(x^2 + x - 3) - 12 = 0 \]

### Step 2: Substitute for Simplicity
Let \(y = x^2 + x\). Then, we can rewrite the equation as:
\[ (y - 2)(y - 3) - 12 = 0 \]
Expanding this gives:
\[ y^2 - 5y + 6 - 12 = 0 \]
which simplifies to:
\[ y^2 - 5y - 6 = 0 \]

### Step 3: Factor the Quadratic
Next, we factor the quadratic equation:
\[ y^2 - 5y - 6 = (y - 6)(y + 1) = 0 \]
This gives us two possible values for \(y\):
\[ y = 6 \quad \text{or} \quad y = -1 \]

### Step 4: Solve for \(x\) from Each \(y\)
Now we will solve for \(x\) using both values of \(y\).

1. **For \(y = 6\)**:
\[ x^2 + x - 6 = 0 \]
Factoring gives:
\[ (x - 2)(x + 3) = 0 \]
Thus, the roots are:
\[ x = 2 \quad \text{and} \quad x = -3 \]

2. **For \(y = -1\)**:
\[ x^2 + x + 1 = 0 \]
The discriminant \(D\) for this equation is:
\[ D = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 \]
Since the discriminant is negative, this quadratic has non-real roots.

### Step 5: Find the Sum of Non-Real Roots
For the quadratic \(x^2 + x + 1 = 0\), the sum of the roots can be found using the formula for the sum of the roots of a quadratic equation, which is given by:
\[ \text{Sum of roots} = -\frac{b}{a} \]
Here, \(a = 1\) and \(b = 1\):
\[ \text{Sum of roots} = -\frac{1}{1} = -1 \]

### Conclusion
Thus, the sum of the non-real roots of the original equation is:
\[ \boxed{-1} \]

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